Question

如圖，我南海某海域A處有一艘捕魚船在作業(yè)時突遇特大風浪，船長馬上向我國漁政搜救中心發(fā)出求救信號，此時一艘漁政船正巡航到捕魚船正西方向的B處，該漁政船收到漁政求救中心指令后前去救援，但兩船之間有大片暗礁，無法直線到達，于是決定馬上調(diào)整方向，先向北偏東60 º方向以每小時30海里的速度航行半小時到達C處，同時捕魚船低速航行到A點的正北1.5海里D處，漁政船航行到點C處時測得點D在南偏東53 º方向上．

(1)求CD兩點的距離；

(2)漁政船決定再次調(diào)整航向前去救援，若兩船航速不變，并且在點E處相會合，求∠ECD的正弦值．    (參考數(shù)據(jù)：，，)

Answer 1

．解：(1)如圖，過點C作CG⊥AB于點G，DF⊥CG于點F，··················································· 1分

則在Rt△CBG中，由題意知∠CBG=30°，

∴CG=BC==7.5， ···························································································· 2分

∵∠DAG=90°，

∴四邊形ADFG是矩形，

∴GF= AD=1.5 ，

∴CF= CGGF=7.5－1.5=6，·································· 4分

在Rt△CDF中，∠CFD=90º，

∵∠DCF =53°，

∴cos∠DCF= ，

∴(海里)． ·························································································· 7分

答：CD兩點距離為10海里． ······························································································· 8分

（2）如圖，設(shè)漁政船調(diào)整方向后t小時能與捕漁船相會合，

由題意知CE=30t，DE=1.5×2×t=3t， ∠EDC=53°， ······························································· 9分

過點E作EH⊥CD于點H， 則∠EHD=∠CHE=90º，

∴sin∠EDH=，

∴EH=EDsin53°=，····························································································· 11分

∴在Rt△EHC中，sin∠ECD=．

答：sin∠ECD=．············································································································· 12分